CS 61 文件IO操作

CS 61 - Programming Assignment 3

Objective

The purpose of this assignment is to give you more practice with I/O, and with left-shifting, aka

multiplying by 2, and useful 2's complement logic.

High Level Description

Store a number to the memory address specified in your assn 3 template. In your program, load

that number to a register, and display it to the console as a 16-bit two's complement binary (i.e.

display the binary value stored in the register, as a sequence of 16 ascii '1' and '0' characters).

Note: Valid numbers are [#-32768, #32767] (decimal) or [x0000, xFFFF] (hex)

Your Tasks

You do not yet know how to take a multi-digit decimal number from user ascii input and convert it

to binary, so for this assignment you are going to let the assembler do that part for you:

you will use the .FILL pseudo-op to take a literal (decimal or hex, as you wish) and translate it into a

16-bit two's complement binary, which will be stored in the indicated memory location; and then

you will Load that value from memory into a register.

You MUST use the provided assn3.asm template to set this up: it ensures that the number to be

converted is always stored in the same location (the memory address specified in your template)

so we can test your work; make sure you fully understand the code fragment we provide.

At this point, your value will be stored in, say, R1: it is now your job to identify the 1's and 0's from

the number and print them out to the console one by one, from left (the leading bit, aka the

leftmost bit, aka bit 15, aka the most significant bit) to right (the trailing bit, aka the rightmost bit,

aka bit 0, aka the least significant bit).

Important things to consider:

● Recall the difference between a positive number and a negative number in 2's complement

binary: if the most significant bit (MSB) is 0, the number is considered positive (or zero);

if it is 1, the number is negative.

● The BRanch instruction has parameters (n, z, p) which tell it to check whether the LMR (Last

Modified Register) is negative, zero, or positive (or any combination of such tests).

Hint: what can you say about the msb of the LMR if the n branch is taken?

Review the workings of the NZP condition codes and the BR instruction here .

● Once you are done inspecting the MSB and printing the corresponding ascii '0' or '1', how would

you shift the next bit into its place so you could perform the next iteration?

Hint: the answer is in the objectives!

Pseudocode:

for(i = 15 downto 0):

if (msb is a 1):

print a 1

else:

print a 0

shift left

Note on creating LC-3 "control structures"

See here for tips on creating LC-3 versions of the branch and loop control structures you are familiar

with from C++ (Resources -> LC-3 Resources -> LC3 Assembly Language -> Control Structures in LC3)

Expected/ Sample output

In this assignment, your output will simply be the contents of R1, printed out as 16 ascii 1's and 0's,

grouped into packets of 4, separated by spaces (as always, newline terminated, but with NO

terminating space!)

So if the hard coded value was xABCD, your output will be:

1010 1011 1100 1101

(The value stored to memory with .FILL was xABCD)

Note:

There are spaces after the first three "packets" of 4 bits (but no space character at end!)
There is a newline after the output - again, there is NO space before the newline
You must use the memory address specified in your template to hold the value to be output

Your code will obviously be tested with a range of different values: Make sure you test your code

likewise!

Uh...help?

● MSB

○ Stands for Most Significant Bit

■ aka "left most bit" or "leading bit" or bit 15

○ When MSB is 0:

■ Means that the number is Not Negative (Positive or Zero)

○ When MSB is 1:

■ Means that the number is Negative

○ Further Reading

■ https://en.wikipedia.org/wiki/Most_significant_bit

● Left Shifting

Left shifting means that you shift all the bits to the left by 1: so the MSB is lost, and is replaced

by the bit on its right. A 0 is "shifted in" on the right to replace the previous LSB.

4-bit Example:

0101 ; #5

When Left Shifted, with 0 shifted in to LSB:

1010 <---- 0101

1010 ; #10

What happened when we left shifted? How did the number change?

When left shifting, the number gets multiplied by 2? Why 2?

Well, what happens when you shift a decimal number one place to the left? Why?

(Practical differences between decimal and binary numbers are that we don't usually limit

decimal numbers to a specific number of places, nor do we usually pad them with leading zeros).